Arithmetics in Z

naoufelabs

Hello all,

I have a problem to define a set of natural numbers that meet the following equation:

m²+1[itex]\equiv[/itex]0[5]

I have found that a set of this equation is : {2,3,7,8}+k*10, k[itex]\in[/itex]N
i.e: k= {0,1,2,3,...........,n}
example: (2+(0*10)²)+1=5
(3+(1*10)²)+1=13²+1=170

Result: m=[2+k*10; 3+k*10; 7+k*10; 8+k*10]

How can I describe this result logically in mathematics ?

Thanks.

dodo

Hello,
I think you more or less did: m is congruent to either 2, 3, 7 or 8 modulo 10.

Notice that this is equivalent to saying that m is congruent to either 2 or 3 modulo 5, since [itex]2 \equiv 7 \pmod 5[/itex] and [itex]3 \equiv 8 \pmod 5[/itex].

You wanted a solution of the equation [itex]m^2 \equiv -1 \equiv 4 \pmod 5[/itex]; you only need to square each of 0,1,2,3,4 modulo 5, and see which are congruent to 4.

Hope this helps!

RamaWolf

Look at sequence A047221 in OEIS ('Numbers that are congruent to {2, 3} mod 5')

naoufelabs

Hello,
thank you for your reply.
I have found a solution :
m²+1 [itex]\equiv[/itex] 0[5]
m² [itex]\equiv[/itex] -1[5]
m² [itex]\equiv[/itex] 4[5]
m [itex]\equiv[/itex] [itex]\pm[/itex]2[5]
m [itex]\equiv[/itex] 2,3[5]

therefore : m= {2+5n ; 3+5n} for any integer n[itex]\geq[/itex]0